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Suppose you throw a 0.054-kg ball with a speed of 11.5 m/s and at an angle of 31.8° above the horizontal from a building 13.1 m high.

(a) What will be its kinetic energy when it hits the ground?
(b) What will be its speed when it hits the ground?

  • Physics -

    conservation of energy

    mgh + (1/2)m(Vi)² = (1/2)m(Vf)²
    2(gh + (1/2)(Vi)²) = (Vf)²
    square root( 2gh +(Vi)²) = (Vf)
    square root( 256.76 + 132,25) = (Vf)
    square root( 389,01) = (Vf)
    square root( 389,01) = (Vf)
    19.7m/s = Vf

    your kinetic energy is

    (0.5)(0.054)(19.7)² = 10.5 J

    enjoy the answer

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